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emmainna [20.7K]
3 years ago
12

I WILL GIVE BRAINLIEST TO THE FIRST OR BEST ANSWER!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Physics
1 answer:
Zielflug [23.3K]3 years ago
3 0
<span>D.Condensation There
</span>
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Which wave is a body wave?<br> a. P wave<br> b. L wave<br> c. B wave<br> d. T wave
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8 0
4 years ago
Read 2 more answers
A 2.35 kg ball is attached to a ceiling by a3.53 m long string. The height of the room is5.03 m .The acceleration of gravity is
Studentka2010 [4]

Answer:-81.29 J

Explanation:

Given

mass of ball m=2.35 kg

Length of string L=3.53 m

height of Room h=5.03 m

Gravitational Potential Energy is given by

P.E.=mgh

where h=distance between datum and object

here Reference is ceiling

therefore h=-3.53 m

Potential Energy of ball w.r.t ceiling

P.E.=2.35\times 9.8\times (-3.53)=-81.29 J

i.e. 81.29 J of Energy is required to lift a ball of mass 2.35 kg to the ceiling    

3 0
3 years ago
The spring is unstretched at the position x = 0. under the action of a force p, the cart moves from the initial position x1 = -8
hammer [34]
Missing figure and missing details can be found here:
<span>http://d2vlcm61l7u1fs.cloudfront.net/media%2Fdd5%2Fdd5b98eb-b147-41c4-b2c8-ab75a78baf37%2FphpEgdSbC....
</span>
Solution:
(a) The work done by the spring is given by
W= \frac{1}{2} k (\Delta x)^2 &#10;
where k is the elastic constant of the spring and \Delta x is the stretch between the initial and final position. Since x1=-8 in=-0.203 m and x2=5 in=0.127 m, we have
W= \frac{1}{2} \cdot 500 N/m \cdot (0.127m-(-0.203m))^2=27.25 J

(b) The work done by the weight is the product of the component of the weight parallel to the inclined plane and the displacement of the cart:
W_W = -F_{//} (x_2 -x_1)
where  the negative sign is given by the fact that F_{//} points in the opposite direction of the displacement of the cart, and where
F_{//}=m g sin 15^{\circ}=6 kg \cdot 9.81m/s^2 \cdot sin 15^{\circ}=15.2 N
therefore, the work done by the weight is
W_W=-15.2 N \cdot (0.203m-(-0.127m))=-5.02 J

8 0
3 years ago
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